1.35 g of a solid acid HZ, was dissolved in 25mL of distilled water. Titration o

Question

1.35 g of a solid acid HZ, was dissolved in 25mL of distilled water. Titration of this unknown acid required 28.5 mL of 0.65 M NaOH solution. What is the equivalent mass of the unknown acid? HZ(aq) + NaOH(aq) rightarrow NaZ(aq) + H_2O A student titrated a standardized solution of HCl to determine the exact concentration of a prepared Mg(OH)_2 solution. He found that it took 20.5 mL of the base solution to neutralize 25.0 mL of a 0.9989M solution of HCl solution. Write the balanced equation for the reaction between the acid and the base. What color should the solution appear at the end point in this titration? Calculate the Molarity of the Mg(OH)_2 solution.

Explanation / Answer

volume of NaOH = 28.5 mL

molarity of NaOH = 0.65 M

moles of NaOH = Molaity x volume in L

= 0.65 M x 0.0285 L

= 0.0185 moles

from equation, 1 mole NaOH reacts with 1 mol HZ

So, moles of HZ = 0.0185 moles

Equivalanet weight of HZ = Mass of HZ / moles of HZ =1.35 g / 0.0185 moles == 72.87 g/mol

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